Application of integration

APPICATION OF INTEGRATIONAPPICATION OF INTEGRATION
SARDAR VALLABHBHAI PATEL
INSTITAUTE OF TECHNOLOGY
GTU

CONTENTCONTENT
1.Volume by cylindrical shell
2.Volume of solids of revolutions by
washers method

APPLICATIONS OF INTEGRATIONAPPLICATIONS OF INTEGRATION

Volumes by
Cylindrical Shells
APPLICATIONS OF INTEGRATION
In this section, we will learn:
How to apply the method of cylindrical shells
to find out the volume of a solid.

Let’s consider the problem of finding the
volume of the solid obtained by rotating about
the y-axis the region bounded by y = 2x2
- x3
and y = 0.
VOLUMES BY CYLINDRICAL SHELLS

If we slice perpendicular to the y-axis,
we get a washer.
 However, to compute the inner radius and the outer
radius of the washer,
we would have to
solve the cubic
equation y = 2x2
- x3
for x in terms of y.
 That’s not easy.

Fortunately, there is a method—the
method of cylindrical shells—that is
easier to use in such a case.

The figure shows a cylindrical shell
with inner radius r1, outer radius r2,
and height h.
CYLINDRICAL SHELLS METHOD

Its volume V is calculated by subtracting
the volume V1 of the inner cylinder from
the volume of the outer cylinder V2 .

Thus, we have:
2 1
2 2
2 1
2 2
2 1
2 1 2 1
2 1
2 1
( )
( )( )
2 ( )
2
V V V
r h r h
r r h
r r r r h
r r
h r r
π π
π
π
π
= −
= −
= −
= + −
+
= −

Let ∆r = r2 – r1 (thickness of the shell) and
(average radius of the shell).
Then, this formula for the volume of a
cylindrical shell becomes:
2V rh rπ= ∆
Formula 1
( )1
2 12r r r= +

The equation can be remembered as:
V = [circumference] [height] [thickness]
2V rh rπ= ∆

Now, let S be the solid
obtained by rotating
about the y-axis the
region bounded by
y = f(x) [where f(x) ≥ 0],
y = 0, x = a and x = b,
where b > a ≥ 0.

Divide the interval [a, b] into n subintervals
[xi - 1, xi ] of equal width and let be
the midpoint of the i th subinterval.
x∆ix

The rectangle with
base [xi - 1, xi ] and
height is rotated
about the y-axis.
 The result is a
cylindrical shell with
average radius ,
height , and
thickness ∆x.
( )if x
( )if x
ix

Thus, by Formula 1, its volume is
calculated as follows:
(2 )[ ( )]i i iV x f x xπ= ∆

So, an approximation to the volume V of S
is given by the sum of the volumes of
these shells:
1 1
2 ( )
n n
i i i
i i
V V x f x xπ
= =
≈ = ∆∑ ∑

The approximation appears to become better
as n →∞.
However, from the definition of an integral,
we know that:
1
lim 2 ( ) 2 ( )
n b
i i an
i
x f x x x f x dxπ π
→∞
=
∆ =∑ ∫

Thus, the following appears plausible.
 The volume of the solid obtained by rotating
about the y-axis the region under the curve
y = f(x) from a to b, is:
where 0 ≤ a < b
2 ( )
b
a
V xf x dxπ= ∫
Formula 2CYLINDRICAL SHELLS METHOD

The argument using cylindrical shells
makes Formula 2 seem reasonable,
but later we will be able to prove it.

Here’s the best way to remember
the formula.
 Think of a typical shell,
cut and flattened,
with radius x,
circumference 2πx,
height f(x), and
thickness ∆x or dx:
( ){
[ ] {2 ( )
b
a
thicknesscircumference height
x f x dxπ∫ 123

This type of reasoning will be helpful
in other situations—such as when we
rotate about lines other than the y-axis.

Find the volume of the solid obtained by
rotating about the y-axis the region
bounded by y = 2x2
- x3
and y = 0.
Example 1CYLINDRICAL SHELLS METHOD

We see that a typical shell has
radius x, circumference 2πx, and
height f(x) = 2x2
- x3
.

So, by the shell method,
the volume is:
( ) ( )
( )
( )
2
2 3
0
2
3 4
0
24 51 1
2 5 0
32 16
5 5
2 2
2 (2 )
2
2 8
π
π
π
π π
= −
= −
 = − 
= − =
∫
∫
V x x x dx
x x x dx
x x

It can be verified that the shell method
gives the same answer as slicing.
 The figure shows
a computer-generated
picture of the solid
whose volume we
computed in the
example.

Comparing the solution of Example 1 with
the remarks at the beginning of the section,
we see that the cylindrical shells method
is much easier than the washer method
for the problem.
 We did not have to find the coordinates of the local
maximum.
 We did not have to solve the equation of the curve
for x in terms of y.
NOTE

However, in other examples,
the methods learned in Section 6.2
may be easier.
NOTE

Find the volume of the solid obtained
by rotating about the y-axis the region
between y = x and y = x2
.

The region and a typical shell
are shown here.
 We see that the shell has radius x, circumference 2πx,
and height x - x2
.

Thus, the volume of the solid is:
( ) ( )
( )
1
2
0
1
2 3
0
13 4
0
2
2
2
3 4 6
V x x x dx
x x dx
x x
π
π
π
π
= −
= −
 
= − = 
 
∫
∫

As the following example shows,
the shell method works just as well
if we rotate about the x-axis.
 We simply have to draw a diagram to identify
the radius and height of a shell.

Use cylindrical shells to find the volume of
the solid obtained by rotating about the x-axis
the region under the curve from 0 to
1.
 This problem was solved using disks in Example 2
in Section 6.2
y x=

To use shells, we relabel the curve
as x = y2
.
 For rotation about
the x-axis, we see that
a typical shell has
radius y, circumference
2πy, and height 1 - y2
.
y x=

So, the volume is:
 In this problem, the disk method was simpler.
( ) ( )
1
2
0
1
3
0
12 4
0
2 1
2 ( )
2
2 4 2
V y y dy
y y dy
y y
π
π
π
π
= −
= −
 
= − = 
 
∫
∫

Find the volume of the solid obtained by
rotating the region bounded by y = x - x2
and y = 0 about the line x = 2.

The figures show the region and a cylindrical
shell formed by rotation about the line x = 2,
which has radius 2 - x, circumference
2π(2 - x), and height x - x2
.

So, the volume of the solid is:
( ) ( )
( )
0
2
1
0
3 2
1
14
3 2
0
2 2
2 3 2
2
4 2
V x x x dx
x x x dx
x
x x
π
π
π
π
= − −
= − +
 
= − + = 
 
∫
∫

Volumes of Solids of
Revolution:
Washer Method

Washers
Consider the area between two functions
rotated about the axis f(x)
a b
g(x
)

Washers
rotated about the axis
Now we have a hollow solid
f(x)
a b
g(x
)

Washers
rotated about the axis
Now we have a hollow solid
We will sum the volumes of washers
f(x)
a b
g(x
)

Washers
f(x)
a b
g(x
)[ ] [ ]{ }2 2
( ) ( )
b
a
V f x g x dxπ= −∫
Outer
Function
Inner
Function

Find the volume of the solid formed by revolving the
region bounded by y = √(x) and y = x² over the interval
[0, 1] about the x – axis.

2 2
([ ( )] [ ( )] )
b
a

2 2
([ ( )] [ ( )] )
b
a
( ) ( )∫ −=
1
0
222
dxxxV π

2 2
([ ( )] [ ( )] )
b
a
( ) ( )∫ −=
1
0
222
dxxxV π
( )∫ −=
1
0
4
dxxxV π

2 2
([ ( )] [ ( )] )
b
a
( ) ( )∫ −=
1
0
222
dxxxV π
( )∫ −=
1
0
4
dxxxV π
1
0
52
5
1
2
1






−= xxV π

2 2
([ ( )] [ ( )] )
b
a
( ) ( )∫ −=
1
0
222
dxxxV π
( )∫ −=
1
0
4
dxxxV π
1
0
52
5
1
2
1






−= xxV π
10
3π
=

Application of integration

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